__Puzzle:__

The riddle is Nine IIT students were sitting in a classroom. Their professor wanted them to test. Next day the professor told all of his 9 students that he has 9 hats, The hats either red or black color. He also added that he has at least one hat with red color and the no. of black hats is greater than the no. of red hats. The professor keeps those hats on their heads and ask them tell me how many red and black hats the professor have? Obviously students can not talk to each other or no written communication, or looking into each other eyes; no such stupid options and no tricks.

Professor goes out and comes back after 20 minutes but nobody was able to answer the question. So he gave them 10 more minuets but the result was the same. So he decides to give them final 5 minutes. When he comes everybody was able to answer him correctly.

So what is the answer? and why?

__Puzzle Solution:__

After first interval of 20 minutes :

Lets assume that their is 1 hat of red color and 8 hats of black color. The student with red hat on his head can see all 8 black hats, so he knows that he must be wearing a red hat.

Now we know that after first interval nobody was able to answer the prof that means our assumption is wrong. So there can not be 1 red and 8 black hats.

After second interval of 10 minutes :

Assume that their are 2 hats of red color and 7 hats of black color. The students with red hat on their head can see all 7 black hats and 1 red hat, so they know that they must be wearing a red hat.

Now we know that after second interval nobody was able to answer the prof that means our assumption is again wrong. So there can not be 2 red and 7 black hats.

After third interval of final 5 minutes :

Now assume that their is 3 hats of red color and 6 hats of black color. The students with red hat on their head can see all 6 black hats and 2 red hats, so they know that they must be wearing a red hat.

Now we know that this time everybody was able to answer the prof that means our assumption is right.So there are 3 red hats and 6 black hats.Now as everybody gave the answer so there can be a doubt that only those 3 students know about it how everybody came to know ?

Then here is what i think, the professor gave them FINAL 5 minutes to answer, so other guys will think that the professor expects the answer after 3rd interval (according to prof it must be solved after 3 intervals), so this is the clue for others.

The solution is very simple i do not understand why the three interval required.

Solution:

In the question there is at least one red hat but no one knows how many red hat is there.

Also black hat is greater then red hat.

If there is only one red hat then the guy with red hat must be able to answer the Q when he saw 8 black hat so there must be more then one red hat.

If there are 4 red hat(there can not be more then 4 red hat) the guys with black hat should have answered the Q. Now the possibility is either two or three.

Now consider there are two red hat, I am one of the guy having red hat in this case i can see 7 black hat and one red hat and considering the other guy having red hat is intelligent ( if he is seeing 8 black hat he must be able to answer the Q so he is seeing one red hat on me ) i can say that there are two red hat. But as i am not able to answer the Q i must be seeing 2 red hat and 6 black hat. If i have black hat one the red hat hat guy must be able to answer that there are 2 red hat and 7 black hat, so the only possibilty left is three red hat and the q solved in second.

Your explanation is right!! Every time the professor comes, guys with red hats increase the count and finally they found that there is one more guy with a red hat other than the 2 they can see and it has to be himself.

Solution:

Case 1. 1 Red Hat 8 Black Hat

The guy with Red Hat can answer it in second.

Case 2. 4 Red Hat 5 Black Hat

Students with Black Hat can answer in second.

Case 3. 2 Red Hat 7 Black Hat

Students with Red Hat can answer in minute.

Students with Red Hat will wait for second to see whether the other student with Red Hat is coming with the answer (1,8) or not if he does not then he must be seeing a Red Hat.

Case 4. 3 Red Hat 6 Black Hat

Students with Red Hat can answer in minute

Students with Red Hat will wait for minute to see whether the other students with Red Hat is coming with the answer (2,7)(as above) or not if he does not then they must be seeing two Red Hat. Wich mean there are three Red Hat.

there are 4 possibilities.

case1: 1 red 8 black

case2: 2 red 7 black

case3: 3 red 6 black

case4: 4 red 5 black

in the first round no one gave the answer:

everybody in the class have 3 conclusions now.

conclusion 1: case 1 Rejected

reason: there is no one in the class who is seeing 8 black hats.

conclusion 2: case 4 Rejected.

reason: there is no one in the class who is seeing 4 red hats.

conclusion 3: only possibility is either case 2 or case 3 is true.

in the second round again there is no answer:

everybody in the class have this conclusion now,

conclusion 1: case 3 Rejected

reason: there is no one in the class who is seeing 3 red hats.

so after 2 rounds it is pretty clear that we are left with only 1 case

which is case 2

there are 2 red hats and 7 Black hats.

so in the third round everybody has the answer.

well i am from NIT.

sorry for previous wrong answer.

This puzzle is tricky actually, and has no correct answer:

this is an extremely beautiful example of deadlock

there are 4 possibilities.

case 1 : 1 Red 8 Black

case 2: 2 Red 7 Black

case 3 : 3 Red 6 Black

case 4 : 4 Red 5 Black

after round 1 and no answer we have 3 conclusion:

1. case1: rejected

reason: no one in room is seeing 8 black hats

2. case4: rejected

reason: No one in the room is seeing 4 red hats

3. either case 2 or case 3 is true.

but after round 2 and no answer we have 3 conclusions again:

1. case 2: rejected (2 red and 7 black)

reason: no one in the room is seeing 7 black hats, if case 2 would be true then red hat wearing students should have get the answer.

(so red hat wearing students will conclude that case 3 is true)

2. case 3: rejected (3 red and 6 black)

reason: no one in the room is seeing 3 red hats, if case 3 would be true then black hat wearing students should have get the answer.

(so black hat wearing students will conclude that case 2 is true)

so in round 3 everyone will be having answer. But 2 different answers

red hat wearing students will say that: there are 3 red and 6 black hats and

black hat wearing students will say that: there are 2 red and 7 black hats.

and both sides have concluded correctly.

so the conclusion is this is a hypothetical situation. and there is no correct answer.

it’s a Deadlock.

please share the correct answer

Answer is simple

2 R 7 B Hat

Consider A and B is sitting and are with red hat.

A can see 7 Black hat and one red hat

B can see 7 Black hat and one red hat

If A is having black hat B can tell the answer that there are 8 Black hat and one red hat.

Same with A.

So both of them will wait for some time and answer the question.

This is an extension of Binayak answer

The solution is very simple i do not understand why the three interval required.

Solution:

In the question there is at least one red hat but no one knows how many red hat is there.

Also black hat is greater then red hat.

If there is only one red hat then the guy with red hat must be able to answer the Q when he saw 8 black hat so there must be more then one red hat.

If there are 4 red hat(there can not be more then 4 red hat) the guys with black hat should have answered the Q. Now the possibility is either two or three.

Now consider there are two red hat, I am one of the guy having red hat in this case i can see 7 black hat and one red hat and considering the other guy having red hat is intelligent ( if he is seeing 8 black hat he must be able to answer the Q so he is seeing one red hat on me ) i can say that there are two red hat. But as i am not able to answer the Q i must be seeing 2 red hat and 6 black hat. If i have black hat the red hat guy must be able to answer that there are 2 red hat and 7 black hat, so the only possibilty left is three red hat and the q solved in second.

( the above and below consequences can thought simultaneously )

now at the same time if the black hat guys saw 3 red hat then they will think as “i am seeing 3 red hat and 5 black hat if i had a red hat then the black hat guys should have answered as 4 red and 5 blacks , since they are silent means i must be wearing black ” so the problem will get solved in the second round itself.

As Randheer singh’s cases:

case 1: can only be solved by a red hat student

case 2: can only be solved by either of red hat students( if case 1 does not hold )

case 3: can be solved by all black hat students( if case 4 does not hold )

case 4: can be solved by all of the black hat students

And one more correction for Randheer singh’s post

{

Scenario I. case 2: rejected (2 red and 7 black)

reason: no one in the room is seeing 7 black hats, if case 2 would be true then red hat wearing students should have get the answer.

(so red hat wearing students will conclude that case 3 is true)

/* this conclusion happens when 3 red hat and 6 black hat and upon failure of your case 2 */

Scenario II. case 3: rejected (3 red and 6 black)

reason: no one in the room is seeing 3 red hats, if case 3 would be true then black hat wearing students should have get the answer.

(so black hat wearing students will conclude that case 2 is true)

/* this conclusion happens when 2 red hat and 7 black hat and upon failure of case3 */

***** since either of the your scenario (I or II ) will satisfy . Your conclusions (within bracket) within scenario’s are mutually independent events they doesn’t hold at any given time *****

}

Why do guys won’t think the otherways

start with 5 black 4 red case instead of 8 black and 1 red case.

Even the case which i mentioned also 5 people will be able to see 4b and 4r case and those all guys knows and since no one answers in first 20 mins this case will be eliminated.

Also people can eliminate the 8 black and 1 red case as well. So totally 2 eliminated in first 20 mins.

So now consider 6b and 3r case. since guys know that 5b4r and 8b1r cases are eliminated. Now in second interval people who have black hat get to know that they are wearing black instead of red as 5b1r is eliminated. Same case they can go with 7b1r as well.

Finally i say that students will not be sure of 6b3r or 7b2r and it depends on whether every one is in synch or not.

suprb

nice

How do each one can see others hat when they are not allowed to see.

You all shitholes don’t have brains and especially eyes.

and that professor who ask question, he asked very smart question but he might have pickup from someone else.

coz that asshole don’t know what he is asking.

Although you all dumbasses counted the correct count of RED and BLACK hats, but did not gave the correct answer.

Question was :

me how many red and black hats the professor have?

Correct Answer: 0 (Zero)

you morons “The professor keeps those hats on their heads”

and asks ” how many hats he have? ”

now please don’t tell me that some of you are from IIT or took high education , coz you are simply not eligible for that.